Question

Two curves $C_1: y=x^2-3$ and $C_2: y=k x^2, k \in R$ intersect each other at two different points. The tangent drawn to $C _2$ at one of the points of intersection $A \equiv\left( a , y _1\right),( a >0)$ meets $C _1$ again at $B \left(1, y _2\right)$ $\left( y _1 \neq y _2\right)$. The value of ' $a$ ' is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

$ \text { Point } A \left( a , y _1\right) \text { lies on } C _1 \text { and } C _2$
$\text { hence } y _1= a ^2-3 \text { and } y _2= ka ^2$
$\Rightarrow a ^2-3= ka ^2$
$\text { now } y = kx ^2 \Rightarrow \frac{ dy }{ dx }=2 kx $
$\left.\therefore \frac{ dy }{ dx }\right]_{\left( a , y _1\right)}=2 ka =\frac{ y _2- y _1}{1- a }\left(\text { But } y _2=1-3=-2\right) $
$=\frac{-2-\left( a ^2-3\right)}{1- a } \Rightarrow 2 ka =\frac{1- a ^2}{1- a }=1+ a $
$2 ka =1+ a$
Substituting $k=\frac{a^2-3}{a^2}$ from (1) in (2) we get $\frac{2 a\left(a^2-3\right)}{a^2}=1+a \Rightarrow 2 a^2-6=a+a^2$ $\Rightarrow a^2-a-6=0 \Rightarrow a=+3, a=-2$ (rejected) ]